Center of mass: 2 objects in triangle system

AI Thread Summary
The discussion revolves around calculating the center of mass for a system of three objects with different masses positioned at the corners of a triangle. The distances provided are a = 4.21 m, b = 1.55 m, and c = 11.93 m. Initial calculations led to incorrect values, with one participant suggesting 5.24 m and another correcting it to 5.5 m. The correct formula involves using the masses and their respective positions to find the center of mass accurately. The conversation emphasizes the importance of proper reference points and calculations in determining the center of mass.
IDKPhysics101
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Three objects of different masses are located at the corners of the triangle shown in the picture. If distance a equals 4.21 m, b equals 1.55 m, and c equals 11.93 m, then what is the center of the mass (in m) of the system in the horizontal with respect to the origin?

X center of mass=6.2/6m+4.21(if set beginning of b as reference point zero then add it in the end to find respect to the origin)=5.24?


I don't have time to type my work right now. But will show work if answer is wrong to see where i went wrong.
 

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IDKPhysics101 said:
Three objects of different masses are located at the corners of the triangle shown in the picture. If distance a equals 4.21 m, b equals 1.55 m, and c equals 11.93 m, then what is the center of the mass (in m) of the system in the horizontal with respect to the origin?

X center of mass=6.2/6m + …

no, not 6.2, try again :smile:
 
well my total answer was 5.24, is that correct?
 
No, neither is 5.24
 
xcm= (2m*4.21) + (3m*5.76) +(m*7.31)
---------------------------------
2m + 3m + m

xcm=5.5?
 
IDKPhysics101 said:
xcm= (2m*4.21) + (3m*5.76) +(m*7.31)
---------------------------------
2m + 3m + m

xcm=5.5?

Yes, 5.5 is correct, but you really do need to go back to your original method (which was better) to see why the 6.2 was wrong :wink:
 
yeah, i realized that i was making 4.21 the origin or the point of reference or something to get the center of mass for just the system no regarding the origin and just adding it in at the end
 
Well that's correct basically and the easier way. We just don't know how you got your 6.2 in the first post.
 
Xcm=(2m*0)+(3m*1.55)+(m*1.55)
-----------------------------
(2m+3m+m)
 
  • #10
Are the 3m and the m mass at the same value of x? At least that is what you wrote there. Look at the sketch again.
 
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